[Numpy-discussion] inplace operations

[Christopher Barker]

BBands wrote:
> If I have a NumPy array like so:
> 
> [[1, 12],
>  [2, 13],
>  [3, 14],
>  [4, 15],
>  [5, 16],
>  [6, 15],
>  [7, 14]]
> 
> How can I do an inplace diff, ending up with this?
> 
> [[1, 0],
>  [2, 1],
>  [3, 1],
>  [4, 1],
>  [5, 1],
>  [6, -1],
>  [7, -1]]

 >>> import numpy as N
 >>> a = N.array([[1, 12],
...  [2, 13],
...  [3, 14],
...  [4, 15],
...  [5, 16],
...  [6, 15],
...  [7, 14]])
 >>>
 >>> a
array([[ 1, 12],
        [ 2, 13],
        [ 3, 14],
        [ 4, 15],
        [ 5, 16],
        [ 6, 15],
        [ 7, 14]])
 >>> a[1:,1] = a[1:,1] - a[:-1,1]
 >>> a
array([[ 1, 12],
        [ 2,  1],
        [ 3,  1],
        [ 4,  1],
        [ 5,  1],
        [ 6, -1],
        [ 7, -1]])
 >>> a[0,1] = 0


 > Also, can I covert to natural logs in place?

 >>> a
array([[  1.,  12.],
        [  2.,  13.],
        [  3.,  14.],
        [  4.,  15.],
        [  5.,  16.],
        [  6.,  15.],
        [  7.,  14.]])
 >>> N.log(a[:,1], a[:,1])
array([ 2.48490665,  2.56494936,  2.63905733,  2.7080502 ,  2.77258872,
         2.7080502 ,  2.63905733])
 >>> a
array([[ 1.        ,  2.48490665],
        [ 2.        ,  2.56494936],
        [ 3.        ,  2.63905733],
        [ 4.        ,  2.7080502 ],
        [ 5.        ,  2.77258872],
        [ 6.        ,  2.7080502 ],
        [ 7.        ,  2.63905733]])

All ufuncs (like log() ) take an optional parameter that is the array 
you want the output to go to. If you pass in the same array, the 
operation happens in place.

By the way, it looks like your first column is an index number. As The 
array structure keeps track of that for you, you might just as well 
store just the data in a single (N,) shaped array.

-Chris


-- 
Christopher Barker, Ph.D.
Oceanographer

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